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The potential energy of a simple harmonic oscillator when the particle is half way to its end point is -
A
$\dfrac{2}{3}E$
B
$\dfrac{1}{8}E$
C
$\dfrac{1}{4}E$
D
$\dfrac{1}{2}E$
Detailed Solution
Total energy of the oscillator: $E = \dfrac{1}{2}ka^2$, where a is the amplitude.
Half way to the end point, $x = \dfrac{a}{2}$.
$PE = \dfrac{1}{2}kx^2 = \dfrac{1}{2}k\left(\dfrac{a}{2}\right)^2 = \dfrac{1}{4} \times \dfrac{1}{2}ka^2$
$PE = \dfrac{E}{4}$
(The remaining $\dfrac{3}{4}E$ is kinetic energy at that point.)
Half way to the end point, $x = \dfrac{a}{2}$.
$PE = \dfrac{1}{2}kx^2 = \dfrac{1}{2}k\left(\dfrac{a}{2}\right)^2 = \dfrac{1}{4} \times \dfrac{1}{2}ka^2$
$PE = \dfrac{E}{4}$
(The remaining $\dfrac{3}{4}E$ is kinetic energy at that point.)